What is the sum from n=0 to infinity of n/(2^{n-1)} ?

The sum from n=0 to infinity of n/(2^{n-1)} is converges

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sum from n=0 to infinity of n/(2^{n-1)}

n = 0 \sum \infty \frac{n}{2 ^{n - 1}}

converges

Solution steps

n = 0 \sum \infty \frac{n}{2 ^{n - 1}}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{1}{n ( 1 + n )}

n = 0 \sum \infty \frac{3}{n ^{2} + 3}

n = 0 \sum \infty \frac{3}{n ^{2} + 4}

n = 0 \sum \infty 12 n (1 - n)^{2}

n = 4 \sum \infty \frac{n}{n ^{2} - 5 n + 6}

What is the sum from n=0 to infinity of n/(2^{n-1)} ?
The sum from n=0 to infinity of n/(2^{n-1)} is converges